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Homotopic Jordan curves in C [pic included]

  1. Mar 6, 2012 #1
    This isn't homework but I'm having trouble understanding the concept of non-homotopic and homotopic Jordan curves.

    My understanding of Jordan curves and homotopy:

    A Jordan curve is a simple closed curve (ie a closed curve that only intersects at the endpoints; f(z1)=f(z2) => z1=z2) such that its exterior is a disjoint union of two open sets:

    i) the "inside" (of the J. curve) is a bounded connected open set
    ii) the "outside" (of the J. curve) is an unbounded connected open set
    iii) the said Jordan curve is the boundary of BOTH "inside" and "outside" (not true for general curves)

    Two closed curves, f1, f2 are homotopic in a connected open set A in C when we can continuously deform with closed curves fk; k in [0,1]. ​


    This is an example from my lecture notes:

    Let A = C\<0>; ie. A is the complex plane with a "hole" at the origin

    There are precisely 3 non-homotopic Jordan curves in A.

    fm4zuq.jpg

    Here's my confusion:
    thanks Office Shredder
     
    Last edited: Mar 6, 2012
  2. jcsd
  3. Mar 6, 2012 #2

    Office_Shredder

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    The boundary of a set is the points you get in it's closure, if the origin is omitted from your topological space to begin with, it cannot possibly be in the closure of any set
     
  4. Mar 6, 2012 #3
    Looks like I had a brain fart there haha, jeez I feel dumb. Thanks.

    Also, I have another question if you could answer it: what makes a Jordan curve unique?

    For example, there is this question: Sketch infinitely many non-homotopic Jordan curves in C\<-1, i, 1>.

    4v1c2v.jpg

    I have these two curves f1 and f2: they have the same index at each point but aren't these two curves essentially the same?
     
  5. Mar 10, 2012 #4
    They're not essentially the same (that is, they're not homotopic) because you can't continuously transform one into the other while staying in the set. That's what homotopic means. Remember that (-1,1,i) have been removed from the set. There's no way you can continuously morph the first curve into the second without having an intermediate curve that passes through i. That's not allowed since i is not in the set. Thus, they're not homotopic.

    Think of the removed points as posts in the ground and the jordan curve as a loop of string. It's not a perfect analogy since technically you would need a string that could pass through itself, but it's still helpful for visualizing these problems. For the infinite number of non-homotopic curves, imagine dropping your loop of string around one post (-1), wrapping the loop and arbitrary number of times arouND -1 and i (in such a way that it doesn't cross itself), and then finally dropping the other end over the final post (+1).

    Good luck on the midterm on Monday :biggrin:
     
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